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NDA 2016 Maths · Mathematics · Matrices · medium

If A=beginpmatrix3&1\\-1&2endpmatrix, then A^2-5A is equal to

Single correct option

If A=(31−12)A=\begin{pmatrix}3&1\\-1&2\end{pmatrix}, then A2−5AA^2-5A is equal to
  1. A. 7I7I
  2. B. −7I-7I
  3. C. OO
  4. D. −5I-5I
Show answer and solution

Answer: (B) −7I-7I

Since tr⁡A=5\operatorname{tr}A=5 and ∣A∣=7|A|=7, the Cayley-Hamilton theorem gives A2−5A+7I=OA^2-5A+7I=O, so A2−5A=−7IA^2-5A=-7I. Direct computation gives A2=(85−53)A^2=\begin{pmatrix}8&5\\-5&3\end{pmatrix}, and A2−5A=(−700−7)A^2-5A=\begin{pmatrix}-7&0\\0&-7\end{pmatrix}.
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Original question written in the NDA 2016 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake